📚 A-Level AQA Chemistry A2 Physical Chemistry Unit 3: Topic Test Revision | A-Level AQA 化学 A2 物理化学单元 3：主题测验复习
This revision guide covers the core content of OxfordAQA International A-Level Chemistry A2 Unit 3 (Physical Chemistry). The topic test typically examines thermodynamics, rate equations, equilibrium constants, acid–base equilibria and electrode potentials. For each area, you need both conceptual clarity and confident algebraic manipulation.
本复习指南覆盖牛津AQA国际A-Level化学A2单元3（物理化学）的核心内容。主题测验通常考察热力学、速率方程、平衡常数、酸碱平衡和电极电势。每个领域既需要清晰的概念理解，也需要熟练的代数运算能力。
1. Thermodynamics: Enthalpy & Entropy | 热力学：焓变与熵变
Thermodynamics in A2 deals with entropy (S) and enthalpy (H) as competing factors that determine whether a reaction is feasible. The total entropy change of the universe is the true criterion for spontaneity.
A2热力学处理熵（S）和焓（H）这两个相互竞争的因素，它们共同决定反应是否可行。宇宙的总熵变是自发性的真正判据。
The entropy change of the system, ΔSsystem, is calculated from the products minus reactants using standard entropy values (units J K⁻¹ mol⁻¹):
系统熵变 ΔSsystem 使用标准熵值（单位 J K⁻¹ mol⁻¹）按产物减反应物计算：
ΔSsystem = ΣS°(products) − ΣS°(reactants)
The entropy change of the surroundings is related to the enthalpy change of the reaction:
环境熵变与反应的焓变相关：
ΔSsurroundings = −ΔH / T
- For an exothermic reaction (ΔH negative), ΔSsurroundings is positive, increasing total entropy.
- 对于放热反应（ΔH 为负），ΔSsurroundings 为正，使总熵增加。
- For an endothermic reaction (ΔH positive), ΔSsurroundings is negative, which may oppose spontaneity.
- 对于吸热反应（ΔH 为正），ΔSsurroundings 为负，可能不利于自发进行。
The total entropy change is the sum of both contributions:
总熵变是两者之和：
ΔStotal = ΔSsystem + ΔSsurroundings = ΔSsystem − ΔH / T
If ΔStotal is positive, the reaction is spontaneous. A common exam trap is forgetting to convert ΔH from kJ to J before dividing by T.
若 ΔStotal 为正，则反应自发。常见考试陷阱是忘记将 ΔH 从 kJ 转换为 J 再除以 T。
2. Gibbs Free Energy | 吉布斯自由能
The Gibbs free energy change combines enthalpy, entropy and temperature into a single, exam-friendly expression:
吉布斯自由能变将焓、熵和温度整合为一个便于考试使用的表达式：
ΔG = ΔH − TΔS
- ΔG < 0: the reaction is spontaneous (feasible).
- ΔG < 0：反应自发（可行）。
- ΔG = 0: the system is at equilibrium.
- ΔG = 0：系统处于平衡状态。
- ΔG > 0: the reaction is not spontaneous as written.
- ΔG > 0：反应按所写方向不自发。
To find the temperature at which a reaction becomes feasible, set ΔG = 0:
求反应变为可行的温度时，令 ΔG = 0：
T = ΔH / ΔS
Worked example | 例题： For CaCO₃ decomposition, ΔH = +178 kJ mol⁻¹ and ΔS = +160 J K⁻¹ mol⁻¹. Calculate the minimum temperature for spontaneous decomposition.
例题： 对于CaCO₃分解，ΔH = +178 kJ mol⁻¹，ΔS = +160 J K⁻¹ mol⁻¹。计算自发分解的最低温度。
ΔG = 0 → T = ΔH / ΔS = 178000 / 160 = 1112.5 K
Above 1112.5 K, ΔG becomes negative and the reaction is feasible. Always note that using 178 instead of 178000 gives a wrong answer by a factor of 1000.
高于1112.5 K时，ΔG变为负值，反应可行。注意若直接使用178而非178000，答案会相差1000倍。
3. Rate Equations | 速率方程
The rate equation expresses the rate of a reaction in terms of reactant concentrations raised to powers called orders:
速率方程将反应速率表示为反应物浓度（带有称为级数的幂次）的函数：
rate = k[A]m[B]n
- k is the rate constant; its units depend on the overall order.
- k 为速率常数，其单位取决于总级数。
- m is the order with respect to A; n is the order with respect to B.
- m 为对 A 的反应级数；n 为对 B 的反应级数。
- The overall order is m + n.
- 总级数为 m + n。
Orders must be determined experimentally — they cannot be predicted from the stoichiometric equation. For a first-order reaction, the half-life is constant:
反应级数必须通过实验测定——不能从化学计量方程式预测。对于一级反应，半衰期恒定：
t½ = ln 2 / k = 0.693 / k
Units of k for common orders:
常见级数下 k 的单位：
| Overall order | Units of k | 总级数 | k 的单位 |
| Zero | 零级 | mol dm⁻³ s⁻¹ | mol dm⁻³ s⁻¹ | mol dm⁻³ s⁻¹ |
| First | 一级 | s⁻¹ | s⁻¹ | s⁻¹ |
| Second | 二级 | mol⁻¹ dm³ s⁻¹ | mol⁻¹ dm³ s⁻¹ | mol⁻¹ dm³ s⁻¹ |
4. Reaction Orders & Rate Constants | 反应级数与速率常数
Three graphical methods allow you to determine the order with respect to a reactant from experimental data.
三种作图方法可从实验数据确定某反应物的反应级数。
- Zero order: A plot of concentration against time is a straight line with a negative gradient; the rate is constant.
- 零级反应： 浓度对时间作图得直线，斜率为负；速率恒定。
- First order: A plot of ln[A] against time is a straight line with gradient −k; the half-life is constant.
- 一级反应： ln[A] 对时间作图得直线，斜率为 −k；半衰期恒定。
- Second order: A plot of 1/[A] against time is a straight line with gradient +k.
- 二级反应： 1/[A] 对时间作图得直线，斜率为 +k。
The initial rates method uses several experiments with varying concentrations. If doubling [A] doubles the rate, the order is 1; if doubling [A] quadruples the rate, the order is 2; if the rate is unchanged, the order is 0.
初始速率法使用多个不同浓度的实验。若浓度[A]加倍使速率加倍，则级数为1；若浓度[A]加倍使速率变为四倍，则级数为2；若速率不变，则级数为0。
Worked example | 例题： For the reaction A + 2B → products, the following initial rates were measured.
例题： 对于反应 A + 2B → 产物，测得以下初始速率。
| Experiment | [A] / mol dm⁻³ | [B] / mol dm⁻³ | Rate / mol dm⁻³ s⁻¹ |
| 1 | 0.10 | 0.10 | 2.0 × 10⁻³ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻³ |
| 3 | 0.10 | 0.30 | 2.0 × 10⁻³ |
Comparing experiments 1 and 2: doubling [A] quadruples the rate, so the order with respect to A is 2. Comparing experiments 1 and 3: tripling [B] does not change the rate, so the order with respect to B is 0. The rate equation is therefore: rate = k[A]².
对比实验1和2：将[A]加倍使速率变为四倍，因此对A的级数为2。对比实验1和3：将[B]增至三倍不影响速率，因此对B的级数为0。速率方程为：rate = k[A]²。
k = rate / [A]² = 2.0 × 10⁻³ / (0.10)² = 0.20 mol⁻¹ dm³ s⁻¹
5. Equilibrium Constants Kc & Kp | 平衡常数 Kc 与 Kp
For a general equilibrium aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration is:
对于一般平衡 aA + bB ⇌ cC + dD，浓度平衡常数为：
Kc = [C]c[D]d / ([A]a[B]b)
- Units of Kc depend on the powers in the expression.
- Kc 的单位取决于表达式中的幂次。
- Only gases and aqueous species appear; pure solids and pure liquids are omitted.
- 只有气体和溶液中的物种出现在表达式中；纯固体和纯液体被省略。
- Kc varies with temperature but not with pressure or concentration.
- Kc 随温度变化，但不随压力或浓度变化。
For gaseous equilibria, Kp is defined using partial pressures. The partial pressure of gas A is:
对于气相平衡，Kp 使用分压定义。气体 A 的分压为：
pA = mole fraction of A × total pressure
Mole fraction = moles of A / total moles of all gases. For the same reaction:
摩尔分数 = A的物质的量 / 所有气体的总物质的量。对于同一反应：
Kp = (pC)c(pD)d / [(pA)a(pB)b]
When calculating Kp, always use equilibrium amounts, not initial amounts. A common error is using moles instead of partial pressures.
计算 Kp 时，务必使用平衡时的物质的量，而非初始量。常见错误为使用物质的量代替分压。
6. Acid–Base Equilibria | 酸碱平衡
The pH scale is defined as:
pH 标度定义为：
pH = −log₁₀[H⁺]
[H⁺] = 10−pH
The ionic product of water at 25 °C is:
25 °C 时水的离子积为：
Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶
For a strong acid, the acid fully dissociates, so [H⁺] = [HA]. For a weak acid, the dissociation is partial and described by Ka:
对于强酸，酸完全解离，因此 [H⁺] = [HA]。对于弱酸，解离不完全，用 Ka 描述：
Ka = [H⁺][A⁻] / [HA]
For a weak acid where the degree of dissociation is small:
对于解离度很小的弱酸：
[H⁺] = √(Ka × [HA])
pH = ½(pKa − log₁₀[HA])
Worked example | 例题： Calculate the pH of 0.050 mol dm⁻³ ethanoic acid, Ka = 1.74 × 10⁻⁵ mol dm⁻³.
例题： 计算 0.050 mol dm⁻³ 乙酸的 pH，Ka = 1.74 × 10⁻⁵ mol dm⁻³。
[H⁺] = √(1.74 × 10⁻⁵ × 0.050) = √(8.7 × 10⁻⁷) = 9.33 × 10⁻⁴ mol dm⁻³
pH = −log₁₀(9.33 × 10⁻⁴) = 3.03
7. Buffer Solutions & pH Curves | 缓冲溶液与 pH 曲线
A buffer solution resists changes in pH when small amounts of acid or base are added. An acidic buffer consists of a weak acid and its conjugate base salt.
缓冲溶液能在加入少量酸或碱时抵抗 pH 变化。酸性缓冲溶液由弱酸及其共轭碱盐组成。
The Henderson–Hasselbalch equation gives the pH of a buffer:
Henderson–Hasselbalch 方程给出缓冲溶液的 pH：
pH = pKa + log₁₀([A⁻] / [HA])
Equivalently, from the Ka expression:
等价地，从 Ka 表达式出发：
[H⁺] = Ka × [HA] / [A⁻]
- Adding small amounts of acid: the conjugate base A⁻ neutralises H⁺.
- 加入少量酸：共轭碱 A⁻ 中和 H⁺。
- Adding small amounts of base: the weak acid HA neutralises OH⁻.
- 加入少量碱：弱酸 HA 中和 OH⁻。
For pH curves:
– Strong acid–strong base: equivalence point at pH 7.
– Weak acid–strong base: equivalence point above pH 7.
– Strong acid–weak base: equivalence point below pH 7.
The buffer region appears as a relatively flat section of the curve before the steep vertical rise.
对于 pH 曲线：
– 强酸-强碱：等当点 pH 为 7。
– 弱酸-强碱：等当点 pH 高于 7。
– 强酸-弱碱：等当点 pH 低于 7。
缓冲区域表现为曲线上陡峭垂直上升前相对平坦的部分。
An appropriate indicator has pKa close to the pH at the equivalence point, and its colour change interval overlaps the vertical section of the curve.
合适的指示剂其 pKa 应接近等当点 pH，且变色范围与曲线的垂直部分重叠。
8. Electrode Potentials & Electrochemical Cells | 电极电势与电化学电池
Standard electrode potentials (E° values) are measured relative to the standard hydrogen electrode (SHE) under standard conditions: 298 K, 1 atm, and 1 mol dm⁻³ solutions.
标准电极电势（E° 值）是在标准条件下相对于标准氢电极（SHE）测定的：298 K、1 atm 和 1 mol dm⁻³ 溶液。
For a cell combining two half-cells:
对于组合两个半电池的电池：
E°cell = E°reduction (cathode) − E°reduction (anode)
The more positive E° value corresponds to the cathode where reduction occurs; the more negative value corresponds to the anode where oxidation occurs. Electrons flow from the more negative electrode to the more positive electrode through the external circuit.
更正 E° 值对应发生还原的阴极；更负的值对应发生氧化的阳极。电子通过外电路从较负的电极流向较正的电极。
The Nernst equation relates the electrode potential to ion concentration:
Nernst 方程将电极电势与离子浓度联系起来：
E = E° + (0.0592 / n) × log₁₀([oxidised form] / [reduced form])
at 25 °C, where n is the number of electrons transferred. As the concentration of the oxidised form increases, E becomes more positive.
在 25 °C 下，其中 n 为转移电子数。当氧化态浓度增加时，E 变得更正。
Predicting spontaneous reactions: a redox reaction is spontaneous when the species with the more negative E° is the reducing agent (oxidised) and the species with the more positive E° is the oxidising agent (reduced), producing a positive E°cell.
预测自发反应：当 E° 较负的物质作为还原剂（被氧化），E° 较正的物质作为氧化剂（被还原），且 E°cell 为正时，氧化还原反应自发进行。
9. Common Exam Mistakes | 常见失分点
Examiners consistently report the same errors in Unit 3 topic tests. Avoid these traps to secure full marks.
考官在单元3主题测验中反复报告相同的错误。避免这些陷阱以获得满分。
- Unit conversion errors: Failing to convert ΔH (kJ) to J before using ΔG = ΔH − TΔS.
- 单位换算错误： 在使用 ΔG = ΔH − TΔS 前未将 ΔH 从 kJ 换算为 J。
- Ignoring state symbols: Including solids or pure liquids in Kc or Kp expressions.
- 忽略状态符号： 在 Kc 或 Kp 表达式中包含固体或纯液体。
- Confusing partial pressure with mole fraction: Partial pressure = mole fraction × total pressure.
- 混淆分压与摩尔分数： 分压 = 摩尔分数 × 总压。
- Assuming weak acids fully dissociate: Weak acids require the approximation [H⁺] = √(Ka[HA]).
- 假设弱酸完全解离： 弱酸需使用近似 [H⁺] = √(Ka[HA])。
- Mixing up anode and cathode: The anode is oxidised (negative in a galvanic cell); the cathode is reduced (positive).
- 混淆阳极和阴极： 阳极发生氧化（原电池中为负极）；阴极发生还原（正极）。
- Rounding too early: Always keep intermediate values to at least 3 significant figures.
- 过早取整： 中间值至少保留3位有效数字。
10. Practice Questions | 自测练习
Work through these questions under timed conditions, then check the solutions.
在限时条件下完成以下问题，然后核对解答。
Question 1 | 问题1： For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹. The entropy change of the system is −199 J K⁻¹ mol⁻¹. Show that the reaction is feasible at 298 K.
问题1： 对于反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH = −92 kJ mol⁻¹。系统熵
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